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[SOLVED] grad operation
Calculate grad(\Phi) in terms of r and r, where \Phi=\frac{1}{r^{3}}
r= xi + yj + zk
r = magnitude (r)
The r cubed term is scalar by the way, it comes out looking bold for some reason.
\Phi = ( \sqrt{x^{2} + y^{2} + z^{2}} )^{-3} = (x^{2} + y^{2} + z^{2})^{\frac{-3}{2}}
grad\Phi = \frac{\partial \Phi}{\partial x}i + \frac{\partial \Phi}{\partial y}j + \frac{\partial \Phi}{\partial z}k
\frac{\partial \Phi}{\partial x} \Rightarrow let (x^{2} + y^{2} + z^{2}) = u
\frac{\partial \Phi}{\partial u} = \frac{-3}{2}u^{\frac{-5}{2}}
\frac{\partial u}{\partial x} = 2x
\frac{\partial \Phi}{\partial x} = 2x(\frac{-3}{2}u^{\frac{-5}{2}})
= -3x((x^{2} + y^{2} + z^{2}))^{\frac{-5}{2}} = -3x( \frac{1}{r^{5}} )
differentials with respect to y and z will be identical except swapping -3x for -3y and -3z.
These will be the terms for i j and k
Giving a final answer of
-\frac{3}{r^{5}} (r).EDIT (again the r^5 term is the magnitude not the vector)
Is this the correct solution? can anybody help?
Homework Statement
Calculate grad(\Phi) in terms of r and r, where \Phi=\frac{1}{r^{3}}
r= xi + yj + zk
r = magnitude (r)
The r cubed term is scalar by the way, it comes out looking bold for some reason.
The Attempt at a Solution
\Phi = ( \sqrt{x^{2} + y^{2} + z^{2}} )^{-3} = (x^{2} + y^{2} + z^{2})^{\frac{-3}{2}}
grad\Phi = \frac{\partial \Phi}{\partial x}i + \frac{\partial \Phi}{\partial y}j + \frac{\partial \Phi}{\partial z}k
\frac{\partial \Phi}{\partial x} \Rightarrow let (x^{2} + y^{2} + z^{2}) = u
\frac{\partial \Phi}{\partial u} = \frac{-3}{2}u^{\frac{-5}{2}}
\frac{\partial u}{\partial x} = 2x
\frac{\partial \Phi}{\partial x} = 2x(\frac{-3}{2}u^{\frac{-5}{2}})
= -3x((x^{2} + y^{2} + z^{2}))^{\frac{-5}{2}} = -3x( \frac{1}{r^{5}} )
differentials with respect to y and z will be identical except swapping -3x for -3y and -3z.
These will be the terms for i j and k
Giving a final answer of
-\frac{3}{r^{5}} (r).EDIT (again the r^5 term is the magnitude not the vector)
Is this the correct solution? can anybody help?